What is the area of a rectangle with vertices at (−6, 3) , (−3, 6) , (1, 2) , and (−2, −1) ?Enter your answer in the box. Do not round any side lengths.What is the area of a triangle with vertices at (0, −2) , (8, −2) , and (9, 1) ?Enter your answer in the box.What is the perimeter of a polygon with vertices at (−2, 1) , (−2, 4) , (2, 7) , (6, 4) , and (6, 1) ?Enter your answer in the box. Do not round any side lengths.What is the perimeter of the rectangle shown on the coordinate plane, to the nearest tenth of a unit?15.3 units20.4 units30.6 units52.0 units
Accepted Solution
A:
(1) the area of a rectangle with vertices at (−6, 3) , (−3, 6) , (1, 2) , and (−2, −1) To find area of rectangle we need to find the length and widthLength = distance between (−6, 3) and (−2, −1) Distance = [tex]\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}[/tex]= [tex]\sqrt{(-2-(-6))^2 + (-1-3)^2}[/tex]= [tex]\sqrt{(4)^2 + (-4)^2}[/tex]=[tex]\sqrt{(16) + 16}[/tex] =[tex]\sqrt{32}[/tex]width = Distance between (−6, 3) , (−3, 6)Distance = [tex]\sqrt{(-3-(-6))^2 + (6-3)^2}[/tex]=[tex]\sqrt{3^2+3^2}= \sqrt{18}[/tex]Area = Length * width = [tex]\sqrt{32} *\sqrt{18} = \sqrt{576}= 24[/tex](2) the area of a triangle with vertices at (0, −2) , (8, −2) , and (9, 1) Area of triangle = [tex]\frac{1}{2} * base * height[/tex]base is the distance between (0,-2) and (8,-2) Distance = [tex]\sqrt{(8-0)^2 + (-2-(-2))^2}[/tex] = 8To find out height we take two vertices (8,-2) and (9,1)Height is the change in y values = 1- (-2) = 3base = 8 and height = 3So area of triangle = [tex]\frac{1}{2} * 8 * 3 = 12[/tex](3) the perimeter of a polygon with vertices at (−2, 1) , (−2, 4) , (2, 7) , (6, 4) , and (6, 1) To find perimeter we add the length of all the sidesDistance between (−2, 1) and (−2, 4) = [tex]\sqrt{(-2+2)^2 + (4-1)^2}[/tex]= 3Distance between(−2, 4) and (2, 7) = [tex]\sqrt{(2+2)^2 + (7-4)^2}[/tex]= 5Distance between (2, 7) and (6, 4) = [tex]\sqrt{(6 - 2)^2 + (4-7)^2}[/tex]= 5Distance between (6, 4) and (6, 1) = [tex]\sqrt{(6 - 6)^2 + (1-4)^2}[/tex]= 3Distance between (6, 1) and (−2, 1) = [tex]\sqrt{(-2-6)^2 + (1-1)^2}[/tex]= 8Perimeter = 3 + 5 + 5 + 3 + 8 = 24 (4) four coordinates are (-7,-1) (-6,4) (3,-3) and (4,2)Length = Distance between (3,-3) and (4,2) = [tex]\sqrt{(4-3)^2 + (2+3)^2}[/tex]= [tex]\sqrt{26}[/tex]Width = Distance between (-6,4) and (4,2) = [tex]\sqrt{(4+6)^2 + (2-4)^2}[/tex]= [tex]\sqrt{104}[/tex]Perimeter = 2(lenght + width) = 2*( [tex]\sqrt{26}[/tex]+[tex]\sqrt{104}[/tex] )= 30.6 units